$(a)\;w(1+2 \alpha \bigtriangleup t)\qquad(b)\;w(1+ \alpha \bigtriangleup t)\qquad(c)\;\large\frac{w}{w(1+2 \alpha \bigtriangleup t)}\qquad(d)\;\large\frac{w}{w(1+ \alpha \bigtriangleup t)}$

Answer : $\large\frac{w}{w(1+2 \alpha \bigtriangleup t)}$

Explanation :

Since the angular momentum is conserved ,

$I_{1}w_{1}=I_{2}w_{2}$

$\large\frac{ml^2}{12}w=\large\frac{m(l+\bigtriangleup l)^2}{12} w_{2}$

$\large\frac{ml^2}{12}w=\large\frac{ml^2}{12}(1+\large\frac{\bigtriangleup l}{l})^2 w_{2}$

$w \approx (1+2 \large\frac{2 \bigtriangleup l}{l})w_{2}$

$[\large\frac{\bigtriangleup l}{l} \to 0]$

$w \approx (1 + 2 \alpha \bigtriangleup t) w_{2} $

$w_{2} \approx \large\frac{w}{(1+2 \alpha \bigtriangleup)}$

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